Question

The questions (i) to (v) refer to the following salt solutions listed A to F:
A. Copper nitrate
B. Iron (II) sulphate
C. Iron (III) chloride
D. Lead nitrate
E. Magnesium sulphate
F. Zinc chloride
(i) Which two solutions will give a white precipitate when treated with dilute hydrochloric acid followed by barium chloride solution?
(ii) Which two solutions will give a white precipitate when treated with dilute nitric acid followed by silver nitrate solution?
(iii) Which solution will give a white precipitate when either dilute hydrochloric acid or dilute sulphuric acid is added to it?
(iv) Which solution becomes a deep/inky blue colour when excess of ammonium hydroxide is added to it?
(v) Which solution gives a white precipitate with excess ammonium hydroxide solution?

Answer 1

(i) Solution of iron (II) sulphate (B) and magnesium sulphate (E) will give a white precipitate on treatment with dilute hydrochloric acid followed by barium chloride solution as per the reactions given below:

FeSO₄ + BaCl₂ → BaSO₄ + FeCl₂

MgSO₄ + BaCl₂ → BaSO₄ + MgCl₂

(ii) Solution of iron (III) chloride (C) and zinc chloride (F) will give a white precipitate on treatment with dilute nitric acid and silver nitrate solution as per the reactions given below:

FeCl₃ + AgNO₃ → 3AgCl + Fe(NO₃)₃
ZnCl₂ + AgNO₃ → 3AgCl + Zn(NO₃)₂

(iii) Solution of lead nitrate (D) gives white precipitate of lead chloride or lead sulphate on treatment with dilute hydrochloric acid or sulphuric acid respectively:

Pb(NO₃)₂ + H₂SO₄ → PbSO₄ + 2HNO₃
Pb(NO₃)₂ + HCl → PbCl₂ + 2HNO₃

(iv) Solution of copper nitrate (A) forms deep blue coloured solution on treatment with excess of ammonium hydroxide solution:

Cu(NO₃)₂ + 2NH₄OH →Cu(OH)₂ + 2NH₄NO₃

(v) Zinc chloride gives a white precipitate on treatment with excess of ammonium hydroxide:

ZnCl₂ + 2NH₄OH → Zn(OH)₂ + 2NH₄Cl